The average selling price of homes in a certain city is $356,300. Assume the variable is normally distributed with a standard deviation of $64,600. If 396 homes are for sale, how many homes will sell for more than $325,000? (Round up to the next whole number.)
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to that Z score. Multiply by 396.
Did I figure this right?
score-mean is 356,300/64,600(SD)= 5.515
My book says that any z values beyond 3.25 us a=0.500.
so I multiplied 0.500 by 396 = 198
I this correct?
To find the number of homes that will sell for more than $325,000, we need to calculate the z-score and then find the corresponding probability.
Step 1: Calculate the z-score
The z-score formula is:
z = (x - μ) / σ
Where:
x = the value we want to convert into a z-score (the selling price of $325,000 in this case)
μ = the population mean (the average selling price of homes in the city, $356,300)
σ = the standard deviation of the population ($64,600)
z = (325,000 - 356,300) / 64,600
z ≈ -0.484
Step 2: Find the probability using the z-score
Now, we need to find the probability of selling a home for more than $325,000. This is equivalent to finding the area under the normal curve to the right of the z-score.
Using a z-table or a calculator, we find that the probability corresponding to a z-score of -0.484 is approximately 0.3146.
Step 3: Calculate the number of homes
To find the number of homes that will sell for more than $325,000, we multiply the probability by the total number of homes for sale.
Number of homes = probability * total number of homes
Number of homes = 0.3146 * 396
Number of homes ≈ 124.58
Since we cannot have a fractional number of homes, we need to round up to the next whole number.
Therefore, approximately 125 homes will sell for more than $325,000.